Mixing
How many sequences strictly growing of length 10 can I write with numbers from 1 to 90?
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I'm trying to solve this combinatorics exercise. My reasoning is very simple so I don't know if it is correct.
Let's start with a simple sequence 1 2 3 4 5 6 7 8 9 10. Then I can think of the other sequences that are created by moving the first digit to the left and adding a new one at the end: 2 3 4 5 6 7 8 9 10 11, 3 4 5 6 7 8 9 10 11 12, etc. So since the multiplication is commutative I can try to calculate the possible sequences by $n(n-1)...(n-9)$ so $90*89*88*87*86*85*84*83*82*81$
combinatorics
asked Jan 25 at 11:48
JackJack
395
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add a comment |
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I'm trying to solve this combinatorics exercise. My reasoning is very simple so I don't know if it is correct.
Let's start with a simple sequence 1 2 3 4 5 6 7 8 9 10. Then I can think of the other sequences that are created by moving the first digit to the left and adding a new one at the end: 2 3 4 5 6 7 8 9 10 11, 3 4 5 6 7 8 9 10 11 12, etc. So since the multiplication is commutative I can try to calculate the possible sequences by $n(n-1)...(n-9)$ so $90*89*88*87*86*85*84*83*82*81$
combinatorics
asked Jan 25 at 11:48
JackJack
395
$endgroup$
add a comment |
$begingroup$
I'm trying to solve this combinatorics exercise. My reasoning is very simple so I don't know if it is correct.
Let's start with a simple sequence 1 2 3 4 5 6 7 8 9 10. Then I can think of the other sequences that are created by moving the first digit to the left and adding a new one at the end: 2 3 4 5 6 7 8 9 10 11, 3 4 5 6 7 8 9 10 11 12, etc. So since the multiplication is commutative I can try to calculate the possible sequences by $n(n-1)...(n-9)$ so $90*89*88*87*86*85*84*83*82*81$
combinatorics
asked Jan 25 at 11:48
JackJack
395
$endgroup$
I'm trying to solve this combinatorics exercise. My reasoning is very simple so I don't know if it is correct.
Let's start with a simple sequence 1 2 3 4 5 6 7 8 9 10. Then I can think of the other sequences that are created by moving the first digit to the left and adding a new one at the end: 2 3 4 5 6 7 8 9 10 11, 3 4 5 6 7 8 9 10 11 12, etc. So since the multiplication is commutative I can try to calculate the possible sequences by $n(n-1)...(n-9)$ so $90*89*88*87*86*85*84*83*82*81$
combinatorics
combinatorics
asked Jan 25 at 11:48
JackJack
395
asked Jan 25 at 11:48
JackJack
395
asked Jan 25 at 11:48
JackJack
395
asked Jan 25 at 11:48
JackJack
395
asked Jan 25 at 11:48
JackJack
395
395
add a comment |
add a comment |
2 Answers
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$begingroup$
Note the following:
- Any subset of $10$ numbers of the $90$ different numbers gives one such sequence (if you order the numbers in the subset increasingly)
- The number of subsets with $10$ elements of the set ${1,2,ldots , 90 }$ is $binom{90}{10}$
So, you get as number of possible strictly increasing sequences of length $10$
$$binom{90}{10} = frac{90cdot 89 cdot ldots cdot 81 }{10!}$$
answered Jan 25 at 11:56
trancelocationtrancelocation
12.7k1827
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It is nothing else than selecting $10$ distinct numbers out of $90$ (to be put in an increasing row afterwards) and there are: $$binom{90}{10}$$ possibilities.
answered Jan 25 at 11:56
drhabdrhab
103k545136
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2 Answers
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active
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2 Answers
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active
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$begingroup$
Note the following:
- Any subset of $10$ numbers of the $90$ different numbers gives one such sequence (if you order the numbers in the subset increasingly)
- The number of subsets with $10$ elements of the set ${1,2,ldots , 90 }$ is $binom{90}{10}$
So, you get as number of possible strictly increasing sequences of length $10$
$$binom{90}{10} = frac{90cdot 89 cdot ldots cdot 81 }{10!}$$
answered Jan 25 at 11:56
trancelocationtrancelocation
12.7k1827
$endgroup$
add a comment |
$begingroup$
Note the following:
- Any subset of $10$ numbers of the $90$ different numbers gives one such sequence (if you order the numbers in the subset increasingly)
- The number of subsets with $10$ elements of the set ${1,2,ldots , 90 }$ is $binom{90}{10}$
So, you get as number of possible strictly increasing sequences of length $10$
$$binom{90}{10} = frac{90cdot 89 cdot ldots cdot 81 }{10!}$$
answered Jan 25 at 11:56
trancelocationtrancelocation
12.7k1827
$endgroup$
add a comment |
$begingroup$
Note the following:
- Any subset of $10$ numbers of the $90$ different numbers gives one such sequence (if you order the numbers in the subset increasingly)
- The number of subsets with $10$ elements of the set ${1,2,ldots , 90 }$ is $binom{90}{10}$
So, you get as number of possible strictly increasing sequences of length $10$
$$binom{90}{10} = frac{90cdot 89 cdot ldots cdot 81 }{10!}$$
answered Jan 25 at 11:56
trancelocationtrancelocation
12.7k1827
$endgroup$
Note the following:
- Any subset of $10$ numbers of the $90$ different numbers gives one such sequence (if you order the numbers in the subset increasingly)
- The number of subsets with $10$ elements of the set ${1,2,ldots , 90 }$ is $binom{90}{10}$
So, you get as number of possible strictly increasing sequences of length $10$
$$binom{90}{10} = frac{90cdot 89 cdot ldots cdot 81 }{10!}$$
answered Jan 25 at 11:56
trancelocationtrancelocation
12.7k1827
edited Jan 25 at 12:15
answered Jan 25 at 11:56
trancelocationtrancelocation
12.7k1827
answered Jan 25 at 11:56
trancelocationtrancelocation
12.7k1827
answered Jan 25 at 11:56
trancelocationtrancelocation
12.7k1827
12.7k1827
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add a comment |
$begingroup$
It is nothing else than selecting $10$ distinct numbers out of $90$ (to be put in an increasing row afterwards) and there are: $$binom{90}{10}$$ possibilities.
answered Jan 25 at 11:56
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
It is nothing else than selecting $10$ distinct numbers out of $90$ (to be put in an increasing row afterwards) and there are: $$binom{90}{10}$$ possibilities.
answered Jan 25 at 11:56
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
It is nothing else than selecting $10$ distinct numbers out of $90$ (to be put in an increasing row afterwards) and there are: $$binom{90}{10}$$ possibilities.
answered Jan 25 at 11:56
drhabdrhab
103k545136
$endgroup$
It is nothing else than selecting $10$ distinct numbers out of $90$ (to be put in an increasing row afterwards) and there are: $$binom{90}{10}$$ possibilities.
answered Jan 25 at 11:56
drhabdrhab
103k545136
edited Jan 27 at 6:56
answered Jan 25 at 11:56
drhabdrhab
103k545136
answered Jan 25 at 11:56
drhabdrhab
103k545136
answered Jan 25 at 11:56
drhabdrhab
103k545136
103k545136
add a comment |
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